TARTARIC ACID 35 found the monohydrate is calculated, 18 parts of water corresponding with 98 of H2S04. Finally, subtraction of the percentage of H2S04 from 100 gives the free S03. (b) In -presence of SOZ: in this case the total acidity must be diminished by that due to the sulphurous acid, that is, before calculating the S03 from the acidity, the number of c.c. of N/io-iodine used in determining the sulphurous acid is divided by 10 and the result subtracted from the number of c.c. of N/2-soda used in the measurement of the. acidity x; from the number thus obtained the S03 is calculated as in (a). The sum of the S03 thus found and of the S02 found directly (i c.c. N/io-iodine = 0-0032 gram of SO 2) is subtracted from 100, the result being the water, from which the H2SO4 is calculated as in the previous case. Lastly, the S03 is calculated by subtracting from 100 the sum of the H2S04 and the S02. Example (of case &) : 3-422 grams of fuming acid, dissolved in water, were made up to 500 c.c. For 100 c.c. of this solution (= 0-6844 gram of substance) 30-10 c.c. of N/a-caustic soda were required, or 5-25 c.c. of N/io-iodine. Then 30-10 0-525 = 29-575, so that the SO3 will be 29'575 X 0-02003 = 0-5924, or, allowing for the amount of the fuming acid in 100 c.c., SO 3 = 86-56%. On the other hand the sulphurous acid will be 5-25 X 0-0032 = 0-0168 i.e., ' S02= 2-46%. Hence the water will be loo (86-56 + 2-46) = 10-98. To 10-98 of water there correspond 59-78 of H2SO4, since 18 : 98 : : 10-98 : 59-78. The free SO3 will therefore be 100 (59-78 + 2'46) = 37'76- The acid thus contains H2S04 5978% S03 37-76% S02 2-46% TARTARIC ACID C4H6Oa = 150 Large, colourless, transparent, odourless, non-hygroscopic crystals of acid taste, readily soluble in water or alcohol. It may be contaminated especially by small quantities of sulphuric acid or sulphates, salts of calcium, potassium, iron, copper, and particularly lead, and sometimes arsenic. It may be adulterated with cream of tartar, potassium and sodium sulphates, alum and oxalic acid. The tests to be made are as follows : 1. Solubility.It should dissolve completely in water (absence of calcium tartrate or sulphate). 1 The reason of this is that, when the acidity is determined in presence of sulphurous acid with methyl orange as indicator, neutrality of the liquid is reached when the acid sulphite, NaHSO3, and not the normal sulphite, Na2SO3, is formed. Consequently, i c.c. of N/io-iodine solution is equivalent; not to o-t c.c., but to only 0-05 c.c. of N-soda and therefore to o-i c.c. N/2-soda.is calculated as percentage, of